LeetCode #1521 — HARD

Find a Value of a Mysterious Function Closest to Target

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

Winston was given the above mysterious function func. He has an integer array arr and an integer target and he wants to find the values l and r that make the value |func(arr, l, r) - target| minimum possible.

Return the minimum possible value of |func(arr, l, r) - target|.

Notice that func should be called with the values l and r where 0 <= l, r < arr.length.

Example 1:

Input: arr = [9,12,3,7,15], target = 5
Output: 2
Explanation: Calling func with all the pairs of [l,r] = [[0,0],[1,1],[2,2],[3,3],[4,4],[0,1],[1,2],[2,3],[3,4],[0,2],[1,3],[2,4],[0,3],[1,4],[0,4]], Winston got the following results [9,12,3,7,15,8,0,3,7,0,0,3,0,0,0]. The value closest to 5 is 7 and 3, thus the minimum difference is 2.

Example 2:

Input: arr = [1000000,1000000,1000000], target = 1
Output: 999999
Explanation: Winston called the func with all possible values of [l,r] and he always got 1000000, thus the min difference is 999999.

Example 3:

Input: arr = [1,2,4,8,16], target = 0
Output: 0

Constraints:

1 <= arr.length <= 10⁵
1 <= arr[i] <= 10⁶
0 <= target <= 10⁷

Step 01

Brute Force Baseline

Problem summary: Winston was given the above mysterious function func. He has an integer array arr and an integer target and he wants to find the values l and r that make the value |func(arr, l, r) - target| minimum possible. Return the minimum possible value of |func(arr, l, r) - target|. Notice that func should be called with the values l and r where 0 <= l, r < arr.length.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Bit Manipulation · Segment Tree

Example 1

[9,12,3,7,15]
5

Example 2

[1000000,1000000,1000000]
1

Example 3

[1,2,4,8,16]
0

Step 02

Core Insight

What unlocks the optimal approach

If the and value of sub-array arr[i...j] is ≥ the and value of the sub-array arr[i...j+1].
For each index i using binary search or ternary search find the index j where |target - AND(arr[i...j])| is minimum, minimize this value with the global answer.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1521: Find a Value of a Mysterious Function Closest to Target
class Solution {
    public int closestToTarget(int[] arr, int target) {
        int ans = Math.abs(arr[0] - target);
        Set<Integer> pre = new HashSet<>();
        pre.add(arr[0]);
        for (int x : arr) {
            Set<Integer> cur = new HashSet<>();
            for (int y : pre) {
                cur.add(x & y);
            }
            cur.add(x);
            for (int y : cur) {
                ans = Math.min(ans, Math.abs(y - target));
            }
            pre = cur;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1521: Find a Value of a Mysterious Function Closest to Target
func closestToTarget(arr []int, target int) int {
	ans := abs(arr[0] - target)
	pre := map[int]bool{arr[0]: true}
	for _, x := range arr {
		cur := map[int]bool{x: true}
		for y := range pre {
			cur[x&y] = true
		}
		for y := range cur {
			ans = min(ans, abs(y-target))
		}
		pre = cur
	}
	return ans
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #1521: Find a Value of a Mysterious Function Closest to Target
class Solution:
    def closestToTarget(self, arr: List[int], target: int) -> int:
        ans = abs(arr[0] - target)
        s = {arr[0]}
        for x in arr:
            s = {x & y for y in s} | {x}
            ans = min(ans, min(abs(y - target) for y in s))
        return ans

// Accepted solution for LeetCode #1521: Find a Value of a Mysterious Function Closest to Target
struct Solution;

use std::collections::HashSet;

impl Solution {
    fn closest_to_target(arr: Vec<i32>, target: i32) -> i32 {
        let mut prev: HashSet<i32> = HashSet::new();
        let n = arr.len();
        let mut res = std::i32::MAX;
        for i in 0..n {
            let mut cur = HashSet::new();
            for x in prev {
                cur.insert(x & arr[i]);
            }
            cur.insert(arr[i]);
            for &x in &cur {
                res = res.min((x - target).abs());
            }
            prev = cur;
        }
        res
    }
}

#[test]
fn test() {
    let arr = vec![9, 12, 3, 7, 15];
    let target = 5;
    let res = 2;
    assert_eq!(Solution::closest_to_target(arr, target), res);
    let arr = vec![1000000, 1000000, 1000000];
    let target = 1;
    let res = 999999;
    assert_eq!(Solution::closest_to_target(arr, target), res);
    let arr = vec![1, 2, 4, 8, 16];
    let target = 0;
    let res = 0;
    assert_eq!(Solution::closest_to_target(arr, target), res);
}

// Accepted solution for LeetCode #1521: Find a Value of a Mysterious Function Closest to Target
function closestToTarget(arr: number[], target: number): number {
    let ans = Math.abs(arr[0] - target);
    let pre = new Set<number>();
    pre.add(arr[0]);
    for (const x of arr) {
        const cur = new Set<number>();
        cur.add(x);
        for (const y of pre) {
            cur.add(x & y);
        }
        for (const y of cur) {
            ans = Math.min(ans, Math.abs(y - target));
        }
        pre = cur;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log M)

Space

O(log M)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.